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Suppose we have a convergent sequence of square integrable holomorphic functions $\{f_n\}$ defined on a domain $D$: they converge to a function $f$. Does $f$ have to be holomorphic too?

This wikipedia article says that $f$ does indeed have to be holomorphic, and that this follows from the fact that for all $f_n$, we have $$\sup\limits_{K}f_n(z)\leq C_K\|f_n\|_{L^2(D)}$$ where $K\subset D$ is any compact subset of the domain.

How does this prove that the limit of the sequence of functions has to be holomorphic?

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    $\begingroup$ I'm guessing what they showed is $f_n \to f$ in $L^2(D)$ implies $f_n \to f$ locally uniformly, and this implies (by general complex analysis theory) that $f$ is holomorphic $\endgroup$ – mathworker21 Apr 13 at 1:47
  • $\begingroup$ The question is very imprecise. In what sense does the sequence converge? If the convergence is pointwise and there is no assumption on the boundedness of $L^{2}$ norms then the claim is false. $\endgroup$ – Kabo Murphy Apr 13 at 5:31
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We can prove your estimate by the mean value property for holomorphic functions:

Suppose $B\subset D$ is a closed disk with center $z_0$ and radius $R$. For all functions $f$ holomorphic in $D$, we have $f(z_0)= \frac{1}{\pi R^2} \int_{|y|\leq R}f(z_0+y)dA(y)$, where $dA$ is 2 dimensional (area) Lebesgue measure. By Cauchy-Schwarz, we get that $|f(z_0)|\leq \frac{1}{\pi R^2} \int_{|y|\leq R}|f(z_0+y)|dA(y)\leq \frac{1}{R\sqrt{\pi}} ||f||_{L^2(B)}\leq \frac{1}{R\sqrt{\pi}} ||f||_{L^2(D)} $

Now let $K\subset D$ compact. By compactness, we can find $B_1, \cdots B_m$ open disks which cover $K$ and whose closures lie in $D$. let r be the minimum radius of the disks. Then by our previous computation, we have shown that $\sup_{z_0 \in K} |f(z_0)| \leq \frac{1}{r\sqrt{\pi}} ||f||_{L^2(D)}$ which proves that for all $K\subset D$ compact, there exists a constant $C_K$ depending only on $K$ and $D$ for which $\sup_{z_0 \in K} |f(z_0)| \leq C_K ||f||_{L^2(D)}$ for all functions $f$ holomorphic in $D$.

It follows that $L^2$ convergence of $\{f_n\}$ implies that $\{f_n\}$ is uniformly Cauchy on each compact subset $K\subset D:$ indeed $f_n-f_m$ is holomorphic hence $\sup_{K} |f_n-f_m| \leq C_K ||f_n-f_m||_{L^2(D)} \to 0$ as $n,m \to \infty$ . In particular $\{f_n\}$ converges locally uniformly, hence has a holomorphic limit by a standard result in complex analysis (one can prove this, for example by Morera's theorem).

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For $|z-a| < R<T$ such that $\{ |z-a|< T\} \subset D$ the Cauchy integral formula yields $$f_n(z) = \frac{1}{2i\pi}\int_{|s-a|= R} \frac{f_n(s)}{s-z}ds= \int_0^1 \frac{f_n(a+R e^{2i\pi t})}{1-zR^{-1} e^{-2i\pi t}}dt \\= \frac{1}{T-R}\int_R^T \int_0^1 \frac{f_n(a+r e^{2i\pi t})}{1-zr^{-1} e^{-2i\pi t}}dt dr$$

If $f_n$ converges in $L^2$ then it converges in $L^1_{loc}$ and hence those integrals and their derivatives and power series (obtained from $\frac{1}{1-zr^{-1} e^{-2i\pi t}} = \sum_{m=0}^\infty z^m r^{-m}e^{-2i\pi mt}$) converge without problems obtaining that $f$ is continuous, $C^\infty$, holomorphic, analytic.

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  • $\begingroup$ OP does not talk about convergence in $L^{2}$ but the Wikipedia article does talk about it. $\endgroup$ – Kabo Murphy Apr 13 at 5:33

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